If A = [[1, sin θ, 1], [-sin θ, 1, sin θ], [-1, -sin θ, 1]]; then for all θ ∈ (3π/4, 5π/4), det(A) l

Q: Why does the middle term of the determinant expansion vanish? It feels like I made a mistake.

It is perfectly natural to feel suspicious when a term vanishes! In the expansion of the determinant along the first row, the middle term involves the minor $\begin{vmatrix} -\sin \theta & \sin \theta \\ -1 & 1 \end{vmatrix}$. Calculating this gives $(-\sin \theta)(1) - (\sin \theta)(-1)$, which is $-\sin \theta + \sin \theta = 0$. The math is robust; the term truly cancels out. Trust your algebraic steps!

Q: I got $\sin \theta \in (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. When I squared it, I got $(\frac{1}{2}, \frac{1}{2})$. Why is the answer $[0, \frac{1}{2})$?

This is the classic 'squaring trap' of JEE! When you square a variable $x$ that lies in an interval containing zero, like $(-a, a)$, the resulting range is $[0, a^2)$. Since $\sin \theta$ passes through zero within your interval, the square must include zero as its minimum value. Never assume the square of an interval is just the square of its endpoints!

Q: My calculated range is $[2, 3)$, but the options are intervals like $(3/2, 3]$. How do I choose?

Great observation! The question asks for an interval that *contains* your result. In set theory, if your set $S = [2, 3)$, you are looking for an option $O$ such that $S \subseteq O$. Since $[2, 3)$ is a subset of $(3/2, 3]$, that option is the correct choice. Always check if your result is a subset of the provided intervals.

Question

If $A = \begin{bmatrix} 1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1 \end{bmatrix}$; then for all $	heta \in (3\pi/4, 5\pi/4)$, $\det(A)$ lies in the interval :

Options:

(A) $(5/2, 4)$

(B) $(3/2, 3]$

(C) $(0, 3/2)$

(D) $(1, 5/2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE journey. Today, we are not just solving a matrix problem; we are peeling back the layers of a mathematical onion.

When you first look at matrix 
$$A = \begin{bmatrix} 1 & \sin 	heta & 1 \ -\sin 	heta & 1 & \sin 	heta \ -1 & -\sin 	heta & 1 \end{bmatrix}$$ 
it might seem like a daunting collection of trigonometric functions. In the world of JEE Advanced, complexity is often a mask for elegance. Our objective is to find the interval of $\det(A)$ for $	heta \in (\frac{3\pi}{4}, \frac{5\pi}{4})$.

### The Art of Expansion

We begin by expanding the determinant along the first row. We take the first element, $1$, and multiply it by its minor, subtract the second element, $\sin 	heta$, multiplied by its minor, and add the third element, $1$, multiplied by its minor.

The expression is:
$$\det(A) = 1 \begin{vmatrix} 1 & \sin 	heta \ -\sin 	heta & 1 \end{vmatrix} - \sin 	heta \begin{vmatrix} -\sin 	heta & \sin 	heta \ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} -\sin 	heta & 1 \ -1 & -\sin 	heta \end{vmatrix}$$

Now, let us perform the simplification:
1. The first minor: $1(1 - (-\sin^2 	heta)) = 1 + \sin^2 	heta$.
2. The second minor: $(-\sin 	heta)(1) - (\sin 	heta)(-1) = -\sin 	heta + \sin 	heta = 0$.
3. The third minor: $(-\sin 	heta)(-\sin 	heta) - (1)(-1) = \sin^2 	heta + 1$.

Combining these, we arrive at:
$$\det(A) = (1 + \sin^2 	heta) - 0 + (1 + \sin^2 	heta) = 2(1 + \sin^2 	heta)$$

### The Trigonometric Trap

Now, we must respect the constraints. We are given $	heta \in (\frac{3\pi}{4}, \frac{5\pi}{4})$.

Visualize the unit circle. At $	heta = \frac{3\pi}{4}$, $\sin 	heta = \frac{1}{\sqrt{2}}$. As we move through the third quadrant towards $\frac{5\pi}{4}$, the sine value decreases, passes through zero at $\pi$, and reaches $-\frac{1}{\sqrt{2}}$ at $\frac{5\pi}{4}$.

Thus, the range of $\sin 	heta$ is $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Since the interval includes zero, the minimum value of $\sin^2 	heta$ is $0$. The maximum value is the square of the magnitude of the endpoints, which is $(\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.

Therefore, $\sin^2 	heta \in [0, \frac{1}{2})$.

### The Final Synthesis

We are at the finish line. We have $\sin^2 	heta \in [0, \frac{1}{2})$. We need to transform this into $2(1 + \sin^2 	heta)$.

First, add $1$ to the inequality:
$$1 + \sin^2 	heta \in [1, \frac{3}{2})$$

Now, multiply the entire range by $2$:
$$2(1 + \sin^2 	heta) \in [2, 3)$$

Our determinant $\det(A)$ lies in the interval **$

Animated Solution for Mathematics - Matrices and Determinants: If $A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1$ ; then for all $θ \in (3 π /4, 5 π /4)$ , $det (A)$ lies in the interval :

Select Answer:

Visualized Solution

Defining Matrix $A$

Expanding $det (A)$

Calculating Minors

Simplifying the Expression

Analyzing the Range of $θ$

Range of $sin θ$

Range of $sin^{2} θ$

Building the Determinant Range

Final Interval and Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Art of Expansion

The Trigonometric Trap

The Final Synthesis

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Visualized Solution

Defining Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)